ECNA2016F

题目大意

给你$nle 100$个数，每次消去一个数的代价是相邻两个数的gcd（循环意义下），最后剩下两个数再取gcd作为代价，问最后消到只剩两个数的代价和最小是多少。

简要题解

dp就好，设f[i][j]表示消去$[i+1,j-1]$里所有数的代价，枚举中间元转移就好，这是区间dp的一般套路嘛，注意取答案的时候只考虑消去n-2个元素，区间长度为n-1，默认最右边的元素留下在加上内部还有一个元素剩下。我真是越学越回去了，天天做煞笔题qaq，不对，哪里天天了。。。

#include <bits/stdc++.h>
using namespace std;
namespace my_header {
#define pb push_back
#define mp make_pair
#define pir pair<int, int>
#define vec vector<int>
#define pc putchar
#define clr(t) memset(t, 0, sizeof t)
#define pse(t, v) memset(t, v, sizeof t)
#define bl puts("")
#define wn(x) wr(x), bl
#define ws(x) wr(x), pc(' ')
    const int INF = 0x3f3f3f3f;
    typedef long long LL;
    typedef double DB;
    inline char gchar() {
        char ret = getchar();
        for(; (ret == '
' || ret == '
' || ret == ' ') && ret != EOF; ret = getchar());
        return ret; }
    template<class T> inline void fr(T &ret, char c = ' ', int flg = 1) {
        for(c = getchar(); (c < '0' || '9' < c) && c != '-'; c = getchar());
        if (c == '-') { flg = -1; c = getchar(); }
        for(ret = 0; '0' <= c && c <= '9'; c = getchar())
            ret = ret * 10 + c - '0';
        ret = ret * flg; }
    inline int fr() { int t; fr(t); return t; }
    template<class T> inline void fr(T&a, T&b) { fr(a), fr(b); }
    template<class T> inline void fr(T&a, T&b, T&c) { fr(a), fr(b), fr(c); }
    template<class T> inline char wr(T a, int b = 10, bool p = 1) {
        return a < 0 ? pc('-'), wr(-a, b, 0) : (a == 0 ? (p ? pc('0') : p) : 
            (wr(a/b, b, 0), pc('0' + a % b)));
    }
    template<class T> inline void wt(T a) { wn(a); }
    template<class T> inline void wt(T a, T b) { ws(a), wn(b); }
    template<class T> inline void wt(T a, T b, T c) { ws(a), ws(b), wn(c); }
    template<class T> inline void wt(T a, T b, T c, T d) { ws(a), ws(b), ws(c), wn(d); }
    template<class T> inline T gcd(T a, T b) {
        return b == 0 ? a : gcd(b, a % b); }
    template<class T> inline T fpw(T b, T i, T _m, T r = 1) {
        for(; i; i >>= 1, b = b * b % _m)
            if(i & 1) r = r * b % _m;
        return r; }
};
using namespace my_header;

const int MAXN = 111 * 2;
int n, a[MAXN], g[MAXN][MAXN], f[MAXN][MAXN];

int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}

int main() {
#ifdef lol
    freopen("F.in", "r", stdin);
    freopen("F.out", "w", stdout);
#endif

    while (scanf("%d", &n) != EOF && n) {
        for (int i = 1; i <= n; ++i) {
            a[i] = a[i + n] = fr();
        }
        n = n << 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j)
                g[i][j] = gcd(a[i], a[j]);
        }
        int ans = INF;
        memset(f, INF, sizeof f);
        for (int i = 1; i <= n; ++i)
            f[i][i + 1] = 0;
        for (int i = 1; i <= n / 2; ++i) {
            for (int j = 1; j + i + 1 <= n; ++j) {
                int k = i + j + 1;
                for (int l = j + 1; l < k; ++l)
                    f[j][k] = min(f[j][l] + f[l][k] + g[j][l] + g[l][k], f[j][k]);
            }
        }
        for (int i = 1; i <= n / 2; ++i) {
            int j = i + n / 2;
            for (int k = i; k <= j; ++k)
                ans = min(ans, f[i][k] + f[k][j] + g[i][k] + g[k][j] + g[k][j]);
        }
        for (int i = 1; i <= n / 2; ++i)
            ans -= g[i][i + 1];
        wt(ans);
    }

    return 0;
}